# Modelling a capacitor with dielectric resistance as a circuit element

The issue of 'how to model the capacitor' arises if the dielectric of a parallel plate capacitor has a finite resistance. As the diagram indicates, current can flow through the capacitor. However, this does not mean that the capacitor acts as a closed circuit. The $r$ is quite large and allows very little current through the dielectric. The current through the dielectric won't enough to prevent a charge build up on the plates. Therefore, the capacitor does not lose its capacitor related properties.

Idea #1 (capacitor with resistance = ideal capacitor and ideal resistor in series):

It might seem intuitive to pull the resistor out of the capacitor and assume that they function as an ideal capacitor and an ideal resistor separately. This model will immediately run into problems because:

1. If you use a DC source, after very long time, the potential drop across the capacitor will be equal to the E.M.F of the source. This would prevent any current from flowing. The potential difference across the capacitor is caused by the charge separation. However, as the dielectric can conduct electricity, the charge on the plates could move. This would never allow the potential difference across the capacitor to be equal to the E.M.F of the source.

Therefore, this model will fail.

Idea #2 (capacitor with resistance = ideal capacitor and ideal resistor in parallel):

After trying Idea #1, the next attempt would be to analyze if considering the dielectric resistance to function as a resistor in parallel. This idea has the following problems:

1. After a sufficiently long time, there would be a constant current in the circuit as the capacitor behaves like an open circuit. There won't be any current passing through the capacitor branch. If you now go back to the original capacitor, there is a path for the charges on the plates of the capacitor to flow. Therefore, the capacitor can never be at a constant potential difference.

How do you model a capacitor which has a finite resistance?

Would I need to go scrap all the capacitance and resistance concepts and start from the fundamentals?

• I don't understand your objection to idea #2 (which is the right way to look at it). Of course the potential across the capacitor will reach a specific value. – Jon Custer Feb 28 '17 at 15:38
• Once the capacitor reaches the specific value, current will stop flowing through it. Now if you go back the original capacitor, as there is a conducting material between the plates, charges must flow through the capacitor to reduce the potential difference. Isn't this a contradiction? – Yashas Feb 28 '17 at 15:40
• @JonCuster I think I figured out now why my argument is invalid. Even if I use the capacitor as it is (conducting dielectric), after sufficiently long time, the capacitor can start to function as a normal resistor. The current entering will not alter the charges on the plates, rather, it will simply pass through the dielectric. The potential drop across this whole component would be the potential drop across the capacitor + the potential drop across the dielectric due to its resistance. – Yashas Feb 28 '17 at 15:48
• (Or the current through the resistor will be determined by the potential across the capacitor). – Jon Custer Feb 28 '17 at 15:48

## 3 Answers

In practical situations all capacitors having dielectric medium (even air) within them have some finite resistance. The system basically behaves like a discharging $RC$ circuit. Let Capacitance $C=\frac{k\epsilon_o A}{d}$

Resistance $R=\frac{\rho d}{A}$

Time Constant of discharge $\tau=RC=\rho k \epsilon_o$

Hence, the discharge current will be $$i=\frac{Q}{RC}e^{\frac{-t}{RC}}=\frac{Q}{\rho k \epsilon_o}e^{\frac{-t}{\rho k \epsilon_o}}$$

This discharge is also called the leakage current.

[$Q$ is the initial charge of capacitor. All other variables which have been used are standard symbols for capacitors]

• "If you use a DC source, after very long time, the potential drop across the capacitor will be equal to the E.M.F of the source. This would prevent any current from flowing. The potential difference across the capacitor is caused by the charge separation. However, as the dielectric can conduct electricity, the charge on the plates could move. This would never allow the potential difference across the capacitor to be equal to the E.M.F of the source" – Yashas Feb 28 '17 at 12:57
• This does not answer the question. – Yashas Feb 28 '17 at 13:30

Instead of guessing a model and trying to verify if it works, let us try to write down the basic equations of electromagnetism and find out which model conforms to the equations.

## Circuit diagram Let $d$ be the distance between the plates of the capacitor.

Let $A$ be the area of the plates of the capacitor.

Let $q$ be the charge on the plates of the capacitor.

Let $r$ be the resistance of the dielectric material of the capacitor.

Let $R$ be the resistance of the external wires.

Let $V$ be the voltage source.

We will not split the capacitor into an ideal capacitor and an ideal resistor (s). We will work out the problem without such simplifications.

## What's happening in the capacitor?

The current through the capacitor can be of two types:

• Displacement current: The current in the capacitor due to the changing electric flux.
• Conduction current: The current which is going through the dielectric.

From conservation of charge, we get: $$I = i_c + i_d \tag{1}$$

where $i_c$ is the conduction current and $i_d$ is the displacement current.

The displacement current appears due to the rate of build up of charge $q$ on the capacitor. It can be shown that:

$$\frac{dq}{dt} = i_d \tag{2}$$  Electric field and potential drop across the capacitor

Assume that the plates are infinitely large. The electric field between the plates of the capacitor can be obtained using Gauss' law.

$$E = \frac{\sigma}{\epsilon} = \frac{q}{A\epsilon}$$

The corresponding potential drop across the capacitor due to the electric field is given by:

$$V_{Ec} = Ed = \frac{qd}{A\epsilon}$$

We can further simplify the equation to

$$V_{Ec} = \frac{q}{C} \tag{3}$$

Apart from the potential drop due to the electric field, the dielectric will have a potential drop due to the conduction current flowing through it:

$$V_{ic} = i_cr \tag{4}$$

As the ends of the dielectric and the ends of the plates coincide, the potential difference across them must be same.

$$V_{Ec} = V_{ic}$$

$$\frac{q}{C} = i_cr \tag{5}$$

## Application of Maxwell–Faraday equation

$$\nabla \times E = -\frac{\partial B}{\partial t}$$

For our problem, we can simplify it to the following form:

$$\int E.dl = -\frac{d\phi_B}{dt} \tag{6}$$

Potential drops and voltage sources:

1. Voltage source $V$
2. Voltage drop due to the capacitor
3. Voltage drop due to the external resistance $R$

Using $(1)$ or $(2)$ and summing up the remaining voltage drops and sources in the circuit, we get:

$$\int E.dl = V - i_cr - IR \tag{7}$$

Using $(2)$ and rearranging $(1)$, we get:

$$i_c = I - i_d = I - \frac{dq}{dt}$$

Substituting the above result in $(7)$, we get:

$$\int E.dl = V - (I - \frac{dq}{dt})r - IR$$

$$\int E.dl = V + r\frac{dq}{dt} - I(R + r)$$

Assuming that the inductance of the wires and the capacitors are negligible and there is no external source of changing magnetic flux, we can equate $(5)$ to zero.

$$V + r\frac{dq}{dt} - I(R + r) = 0 \tag{8}$$



## Analyzing equation $(8)$

After a bit of reverse engineering (guesswork), I discovered that an equation similar to $(8)$ is obtained for the following diagram. We won't solve the above circuit with all steps. I will quickly skim through the steps and establish the differential equation for the above circuit.

$$I = i_d + i_c,\space i_d = \frac{dq}{dt}, \space => i_c = I - \frac{dq}{dt}$$

Applying Kirchhoff's voltage rule for the above circuit, we get:

$$V - i_cR_1 - IR_2 = 0$$

$$V - (I - \frac{dq}{dt})R_1 - IR_2 = 0$$

$$V - R_1\frac{dq}{dt} - I(R_1 + R_2) = 0 \tag{9}$$



Comparing equation $(8)$ with $(9)$

Careful observation reveals that the structure of equation $(9)$ is exactly identical to the equation $(8)$.

We can draw the following analogies between the two equations:

• $r = R_1$
• $R + r = R_1 + R_2$, $=> R + r = r + R_2$m $=> R = R_2$

It can be infered from the above analogies that a capacitor with leakage current behaves as an ideal capacitor with a resistor connected in parallel to it.

## Summary If you consider the resistance of dielectric (touching both plates) in a DC circuit, the equivalent system will be just a resistance (having resistance equal to resistance of dielectric). The capacitor plates are just metal plates. In case the medium can conduct electricity the significance of capacitor is lost.

• Please read the question again. I dedicated an entire paragraph to address the issue you have stated. – Yashas Feb 28 '17 at 13:41
• @YashasSamaga The plates will be charged. I never denied that. Q/kC=E=iR will hold where E is the emf of source. (k is dielectric constant) – 2017 Feb 28 '17 at 13:44
• Potential drop across capacitor plates will be equal to potential drop across resistance.. – 2017 Feb 28 '17 at 13:45
• "As the diagram indicates, current can flow through the capacitor. However, this does not mean that the capacitor acts as a closed circuit. The $r$ is quite large and allows very little current through the dielectric. The current through the dielectric won't enough to prevent a charge build up on the plates. Therefore, the capacitor does not lose its capacitor related properties." – Yashas Feb 28 '17 at 13:46
• Ok, that is a good point. The potential drop across the capacitor will be equal to the potential drop across the resistance. – Yashas Feb 28 '17 at 13:46